Answer:
Maximum height=9.8 ft and Time taken by soccer ball will be 0.8 secs
Step-by-step explanation:
Since, the ball kicked will take the motion of projectile, therefore using the equation: h(t)= \(at^{2} +vt+d\), where h(t) is the height of the soccer ball, a is the acceleration whose value is -16 ft/sec^2 and v= 35 feet and d is the starting height which is equal to zero.
Therefore, h(t)=\(at^{2} +vt+d\) (1)
Differentiating this equation with respect to t,
\(h^{'}(t)\)= \(2at+v\)
\(0\) =\(2(-16)t+25\)
\(32t\)=\(25\)
\(t= 0.781sec\)
Substituting the value of t in equation (1),
\(h(t)=-16(0.781)^{2} +25(0.781)\)
=\((-16)(0.609)+19.525\)
=\(9.766\)
≈9.8ft
Hence, option C is correct.
Now, In order to determine the time taken until the soccer ball hits the ground, we take the equation:
\(h(t)= at^{2} +vt+d\)
Since, when the ball hits the ground, the height will become equal to zero, therefore we have, h(t)=0
Now, \(h(t)= at^{2} +vt+d\)
\(0= -32.174t^{2} +25t\)
\(0=t(-32.174t+25)\)
Then, one solution is t=0 and the other is: \(0=-32.174t+25\)
\(32t=25\)
\(t=\frac{25}{32.174}\)
\(t=0.777 sec\)
\(t\)≈\(0.8 sec\)
Hence, option B is correct.
